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Question Number 156119 by Ghaniy last updated on 08/Oct/21

solve : ((1+2x)/(1+(√(1+2x))))+((1−2x)/(1−(√(1−2x))))=1

solve:1+2x1+1+2x+1−2x1−1−2x=1

Commented by immortel last updated on 08/Oct/21

Commented by Ghaniy last updated on 12/Oct/21

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Answered by MJS_new last updated on 08/Oct/21

defined for −(1/2)≤x<0∨0<x≤(1/2) let t=1+(√(1+2x)) ⇒ t≥1 ⇔ x=((t(t−2))/2) x≠0 ⇒ t≠0∧t≠2 insert ⇒ (((t−1)^2 )/t)−((t^2 −2t−1)/(1−(√(−t^2 +2t+1))))=1 ((t^2 −2t−1)/(1−(√(−t^2 +2t+1))))=−((t^2 −3t+1)/t) ((1−(√(−t^2 +2t+1)))/(t^2 −2t−1))=−(t/(t^2 −3t+1)) (√(−t^2 +2t+1))=−((3t^3 −3t^2 +2t−1)/(t^2 −3t+1)) squaring & transforming t^6 −7t^5 +((35)/2)t^4 −18t^3 +6t^2 =0 t^2 (t−2)^2 (t^2 −3t+(3/2))=0 t≠0∧t≠2∧t≥1 ⇒ t=((3+(√3))/2) ⇒ x=((√3)/4)

definedfor−12⩽x<0∨0<x⩽12lett=1+1+2x⇒t⩾1⇔x=t(t−2)2x≠0⇒t≠0∧t≠2insert⇒(t−1)2t−t2−2t−11−−t2+2t+1=1t2−2t−11−−t2+2t+1=−t2−3t+1t1−−t2+2t+1t2−2t−1=−tt2−3t+1−t2+2t+1=−3t3−3t2+2t−1t2−3t+1squaring&transformingt6−7t5+352t4−18t3+6t2=0t2(t−2)2(t2−3t+32)=0t≠0∧t≠2∧t⩾1⇒t=3+32⇒x=34

Commented by Ghaniy last updated on 12/Oct/21

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